Find the point b/w two charges where the electric field is zero?

Click For Summary
SUMMARY

The discussion centers on determining the point between two positive charges (16 microcoulombs and 4 microcoulombs) where the electric field is zero, given they are separated by 3 meters. Participants clarify that the electric field is zero when the forces exerted by the two charges on a test charge balance out, leading to the equation E1 + E2 = 0. The confusion arises from equating electric fields with forces, which are distinct concepts; the electric field is defined by the force on a test charge, not the forces between the two charges themselves.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with Coulomb's Law
  • Basic algebra for solving equations
  • Concept of test charges in electric fields
NEXT STEPS
  • Study the concept of electric field lines and their implications
  • Learn how to apply Coulomb's Law in different scenarios
  • Explore the superposition principle in electric fields
  • Practice solving problems involving multiple charges and electric fields
USEFUL FOR

Students of physics, educators teaching electromagnetism, and anyone interested in understanding electric fields and forces between charges.

kashan123999
Messages
98
Reaction score
0

Homework Statement



Two positive charges of 16 micro coulomb and 4 micro coulomb are separated by a distance 3 m along a straight line...find the spot on the line joining the two charges where Electric field is zero?

Homework Equations



E = F/Q and F = Q1Q2/R^2

The Attempt at a Solution



AS i tried it i thought electric field is zero when forces b/w two charges will become equal,i took a point P on a line,so that E2 is a distance "d" from P,and E1 is at a distance "3-d" from p...so i applied F1=F2 but the answer came out to be 3 which wasn't actually,i checked the textbook and they equalled E1=E2


actually i know its solution but couldn't grasp the idea of the difference b/w electric field/coloumb's law and electric force? i couldn't help thinking my self it can also to be F1 = F2 but why it is E1 = E2? what is the idea notion behind it? please can anyone explain it in layman's terms?
 
Physics news on Phys.org
Force between two charges is ALWAYS equal and opposite by Newton's 3rd law.
Thats why you are getting the same distance as given in the question.
Electric field is given by the force on a small positive test charge. The force on this charge (or any other charge for that matter) is zero when electric field is zero.
 
Enigman said:
Force between two charges is ALWAYS equal and opposite by Newton's 3rd law.
Thats why you are getting the same distance as given in the question.
Electric field is given by the force on a small positive test charge. The force on this charge (or any other charge for that matter) is zero when electric field is zero.

can you elaborate it a bit more please sir
 
kashan123999 said:
actually i know its solution but couldn't grasp the idea of the difference b/w electric field/coloumb's law and electric force? i couldn't help thinking my self it can also to be F1 = F2 but why it is E1 = E2? what is the idea notion behind it? please can anyone explain it in layman's terms?

You do not want E1=E2 here, you want E1+E2=0.
 
CAF123 said:
You do not want E1=E2 here, you want E1+E2=0.

how?
 
kashan123999 said:
can you elaborate it a bit more please sir
You don't care about the force of the two charges on each other. The field relates to the force that would be experienced by some other charge placed at that point. So if you want to do it by balancing the forces, imagine introducing a 'test' charge q at distance d from one and 3-d from the other and calculate the two forces exerted on it.
 
  • Like
Likes   Reactions: 1 person
kashan123999 said:
how?
The E field from a positive charge is radially outward and proportional to 1/r2. So only for large enough r does E tend to zero. Solving like E1=E2, is saying I want the E field from charge 1 to be zero and I want the E field from charge 2 to be zero and then solving for x. But there is no x.

If you draw a picture and sketch the E field lines, you can see there is a point between the two charges were the total E field is zero, total meaning the superposition of the E fields from both sources.
 
  • Like
Likes   Reactions: 1 person
haruspex said:
You don't care about the force of the two charges on each other. The field relates to the force that would be experienced by some other charge placed at that point. So if you want to do it by balancing the forces, imagine introducing a 'test' charge q at distance d from one and 3-d from the other and calculate the two forces exerted on it.

couldn't get the answer by F1= -F2 or F1 + F2 = 0..please help
 
kashan123999 said:
couldn't get the answer by F1= -F2 or F1 + F2 = 0..please help
The equation is correct, but would you be able to show more of your work?
 
  • #10
(Kq1q2)/(d^2) + (kq1q2)/(3-d)^2 = 0

kq1q2 can be taken as a common factor,so it will be eliminated...

1/d^2 + 1/(3-d)^2 = 0

Taken L.C.M,as RHS = 0 Denominator will become 1,

(3-d)^2 + d^2 = 0

9 + d^2 - 6d + d^2 = 0

2d^2 - 6d + 9 = 0

now how to factor that?
 
  • #11
also sir please i am still not able to grasp the notion of making E1 = E2,couldn't comprehend the question clearly :( please elaborate in the simple possible terms if you can sir
 
  • #12
kashan123999 said:
(Kq1q2)/(d^2) + (kq1q2)/(3-d)^2 = 0

kq1q2 can be taken as a common factor,so it will be eliminated...

What is q1 and what is q2? You want to find the point between the 16μC and the 4μC where the E field is zero. So you do not need to consider forces at all. You can do it with forces however. A test charge placed at the location of zero E field would experience zero net force. There are two forces acting on the test charge: the force from the 16μC charge and the force from the 4μC. We do not need to be concerned with the force between these two charges.

Put the 16μC at the origin for simplicity. The test charge is a distance d from the 16μC and a distance 3-d from the 4μC charge.
What is the force acting on the test charge due only to the 16μC?
What is the force acting on the test charge due only to the 4μC?

Note their directions too. Like repels like and take the test charge to be positive. (it could equally have been negative).
 
  • #13
kashan123999 said:
(Kq1q2)/(d^2) + (kq1q2)/(3-d)^2 = 0
What are q1 and q2? With the forces approach, you have three charges: the two given charges (call those q1 and q2) and a test charge q. You want the force between q1 and q balancing the force between q2 and q.
 
  • #14
haruspex said:
What are q1 and q2? With the forces approach, you have three charges: the two given charges (call those q1 and q2) and a test charge q. You want the force between q1 and q balancing the force between q2 and q.

couldn't get that :(
 
  • #15
kashan123999 said:
couldn't get that :(
An electric field exerts a force on a charge placed within the field. Where the field is zero, the force is zero.
You are asked for a locus at which the net field from two given charges is zero. That means that a 'test' charge (i.e. some third charge) placed at that point will feel no net force. That is, the two forces it feels from the two given charges balance out.
If the two given charges are q1 and q2, placed distance s apart, and a third charge q is placed distance d from q1 and distance s-d from q2:
- what force is exerted on it by q1?
- what force is exerted on it by q2?
- what is the magnitude of the net force?
 
  • Like
Likes   Reactions: 1 person

Similar threads

Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
990
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
35
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 68 ·
3
Replies
68
Views
8K
  • · Replies 12 ·
Replies
12
Views
3K